MAGNETIC FIELDS OF STEADY ELECTRIC CURRENTS, BIOT SAVART LAPLACE LAW, AND ITS APPLICATIONS

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1 MAGNETIC FIELDS OF STEADY ELECTRIC CURRENTS, BIOT SAVART LAPLACE LAW, AND ITS APPLICATIONS In these notes I explain the magnetic fields of steady electric currents. The fields of time-dependent currents are more complicated, and I ll discuss them a few lectures later. Let s start with the simplest case of an infinite straight wire carrying a current I. The magnetic field os such a wire is B = µ I π ˆφ s 1) where µ is the fundamental constant of the MKSA system of units called the vacuum permeability, µ = 1 7 Tm/A, ) s is the distance from the wire, and ˆφ is a unit vector in the circular direction around the wire in the plain to the wire. Specifically, if you are looking down the wire and the current flows away from you, then the circular direction ˆφ of the magnetic field is clockwise, while if the current flows toward you, then the magnetic field is counterclockwise. Here are the pictures of the magnetic field lines for the two cases: The effect of this magnetic field on another long wire parallel to the first wire is an attractive force if the currents in the two wires flow in the same direction, and a repulsive force if the 1

2 currents flow in the opposite directions, I 1 I I 1 I I 1 I F F F F F F 3) Here is the graphical explanation of the force s direction for the currents in the same direction: The magnitude of the force between two wires per unit of wire s length is 4) F L = µ π I 1I d 5) where d is the distance between the wires, while the µ constant of the MKSA system of units is exactly µ = 1 7 Tm/A = 1 7 N/A. 6) In other words, the Ampere the MKSA unit of electric current is defined such that two long parallel wires separated by 1 m distance and each carrying 1 A current are attracted or repelled with a force of 1 7 Newtons per meter of length.

3 In the Gaussian system of units, there is no µ. Instead there are factors 1/c where c is the speed of light in vacuum) all over the place, For example, the Lorentz Force on a particle in Gaussian units is the magnetic force on a current-carrying wire is F = q E + v ) c B, 7) F = 1 c wire the magnetic field of an infinite straight wire is I d l B, 8) B = I c ˆφ s, 9) and the force between parallel wires is F L = c I 1I d. 1) The wires of geometries other than an infinite straight line create magnetic fields much more complicated that 1). For a steady current in a wire of most general geometry, there is an integral formula known as the Biot Savart Laplace equation or Biot Savart Laplace Law: Br) = µ wire I dr r r r r 3 11) in MKSA units, or Br) = 1 c wire Idr r r r r 3 1) ingaussianunits. Intheseformulae, ther spansthewireandthedr = d listheinfinitesimal 3

4 length vector along the wire in the direction of the current. The expression r r r r 3 = unit vector from r to r distance between r and r ) 13) should befamiliar to you fromthe Coulomb Law forthe electric field of a charge distribution, for example the electric field of a charged wire is Er) = 1 ǫ wire r r r r λdl. 14) 3 But there is a crucial difference between the Biot Savart Laplace equation 11) and the Coulomb equation 14) the cross product of Id l with the kernel 13) in the BSL equation 11). In these notes, I shall explore the consequences of this vector product for several examples of wire geometries. Example#1: Infinite Long Wire Formyfirst example, let mereproduce eq. 1)forthemagneticfield ofaninfinite straight wire from the Biot Savart Laplace Law. Let me use the coordinate system where the wire runs along the z axis with the current flowing in the +ẑ direction. Consequently, in the BSL equation Br) = µ wire I dr r r r r 3 11) we have r =,,z ) for a variable z but fixed x = y =, and Idr = +I dz ẑ; on the other hand, the coordinates x,y,z) of the point r where we measure the magnetic field are completely general. Therefore, r r = xˆx + yŷ + z z )ẑ, 15) I dz ẑ r r ) = Idz xŷ yˆx + z z ) ), 16) r r = x + y + z z ), 17) r r 3 = x +y +z z ) ) 3/, 18) 4

5 and plugging all these formulae into eq. 11) gives us Bx,y,z) = Iµ xŷ yˆx) + dz x +y +z z ) ) 3/. 19) To evaluate the integral here, let s change the integration variable from z to α = arctan Consequently, s z z = z = z s tanα = z s ctanα where s = x +y. ) dz = + sdα sin α, 1) x +y +z z ) = s + s tan α = s sin α, ) dz x +y +z z ) ) 3/ = sdα sin α sin3 α s 3 = sinαdα s while the angle α runs from for z to π for z +. Hence, = d cosα) x +y, 3) + dz π x +y +z z ) ) 3/ = d cosα) x +y = cosπ)+cos) x +y = x +y 4) and therefore Bx,y,z) = µ I π = µ I π xŷ yˆx x +y in Cartesian coordinates ˆφ in cylindrical coordinates. s 5) Thus, the basic formula 1) for the infinite straight wire indeed follows from the general Biot Savart Laplace equation.. 5

6 Example#: Circular Ring For the next example, consider a wire shaped into a circular ring of radius R. For simplicity, let me limit the calculation of the magnetic field to the axis of the ring, otherwise we would have to deal with elliptic integrals. Let s use the coordinate system where the ring lies in the xy plane while its symmetry axis is the z axis, thus z r y I r x 6) Along the circular wire, r = Rcosφˆx + Rsinφŷ, 7) dr = R sinφˆx + cosφŷ)dφ, 8) while the points r where we measure the magnetic field are restricted to r = zẑ, hence r r = Rcosφˆx Rsinφŷ + zẑ, 9) sinφˆx + cosφŷ) cosφˆx sinφŷ) = = sinφcosφ ˆx ˆx ŷ ŷ = ) +sin φ ˆx ŷ = +ẑ ) cos φ ŷ ˆx = ẑ ) = sin φ+cos φ)ẑ = ẑ, 3) 6

7 sinφˆx + cosφŷ) ẑ = sinφ ˆx ẑ = ŷ ) + cosφ ŷ ẑ = +ˆx ) = cosφˆx + sinφŷ, 31) ) I dr r r ) = IR zcosφˆx + zsinφŷ + Rẑ dφ, 3) r r = R + z, 33) r r 3 = R +z ) 3/. 34) Plugging all these formulae into the Biot Savart Laplace equation, we obtain B,,z) = µ = µ wire Idr r r r r 3 IR R +z ) 3/ π ) dφ zcosφˆx + zsinφŷ + Rẑ, 35) where the integral evaluates to π ) dφ zcosφˆx + zsinφŷ + Rẑ π π π = zˆx dφ cosφ + zŷ dφ sinφ + Rẑ dφ = zˆx + zŷ + Rẑ π = πrẑ. Altogether, the magnetic filed along the ring s axis is 36) B,,z) = µ I In particular, at the center of the ring, the magnetic field is Bcenter) = µ I R R R +z ẑ. 37) ) 3/ ẑ. 38) Note: on the diagram 6), the current in the wire flows counterclockwise; consequently, the magnetic field 37) points up, in the +ẑ direction. For a clockwise current, we would 7

8 have an opposite sign of I dr and hence opposite direction ẑ of the magnetic field down. This is an example of the right screw rule for the current loops: turn a right screw almost all the screws are right) in the direction of the current in the loop, and the screw will move in the direction of the B field. Equivalently, you may use the right hand rule: curl the fingers of your right hand around the loop in the direction of the current, and your thumb will point the direction of the B field. Segments: In many cases, a wire is made of several segments. Each segment has a simple geometric shape a piece of a straight line, or a circular arc but the overall geometry can be quite elaborate. For example, consider a star made of 5 straight-line segments, 39) For a wire like this, the Biot Savart Laplace integral over the whole wire becomes a sum of integrals over the individual segments, Br) = µ I segments i segment#i dr r r ) r r 3. 4) Let s work out the integrals here for the straight-line and the circular-arc segments, and than we shall see a few interesting combinations. 8

9 Example#3: Straight-Line Segment: Consider a wire segment which follows a straight line from point r 1 to point r 1. Let s picture the triangle made by the two ends of this segment and by the point r where we measure the magnetic field: O r 1 h α 1 41) r r α Since the wire segment is straight, the infinitesimal vector d l = dr along the segment has a fixed direction, same as r r 1. Consequently, the vector product in the numerator of the BSL integral remains constant along the whole segment. Indeed, dr r r ) = dr r R O ) dr r R O ) where the second term vanishes since dr r R O ) r r 1 ) dr r R O ) 4) = d l h, where d l = dr is theinfinitesimal length element along thestraight segment, and h = r R O isthe heightof thetriangle41). In other words, h is theline tothe point rwhere we measure the magnetic field from the wire segment or from the extrapolated straight line of the wire segment in the direction to the segment. Note: if we measure the magnetic field at a point r which happens to lie right on the extrapolated straight line of the wire segment, then h = and hence d l h. Consequently, the whole BSL integral vanishes regardless of the denominator s details, and the magnetic field of the segment is zero. Thus, straight segments pointing directly towards or directly away from r do not contribute to the magnetic field at r. 9

10 For h, the direction of the magnetic field is the direction of the vector product d l h in the numerator of the BSL integral. This direction is to the wire and to the h; in other words, the direction of Br) is to the whole triangle 41). The specific perpendicular obtainsfromtherightscrewrule: Iffromyourpointofview, thecurrentflowsintheclockwise direction around r as it does on figure 41) then take the perpendicular which points away from you. OOH, if you see the current flows counterclockwise around r, then take the perpendicular which points towards you. Now that we know the direction of the magnetic field, let s find its magnitude B = µ I l dl h r r 3 43) In this formula, l is the coordinate along the wire; let s I take it s origin l = to be the point O where the height h of the triangle touches the wire or the extrapolated line of the wire. In terms of this l, l 1 r r = l + h = r r 3 = l +h ) 3/, 44) so the BSL integral 43) becomes B = µ I l h dl l +h ) 3/. 45) l 1 To evaluate this integral, we proceed similarly to eqs. ) through 4): we change the integration variable l to the angle α = arcctan l h = l = h ctanα, 46) hence dl l +h ) 3/ = d cosα) h 47) 1

11 and therefore l l 1 α h dl l +h ) 3/ = d cosα) h α 1 = cosα 1 cosα h 48) where the angles α 1 and α are exactly as shown on the diagram 41). Altogether, the magnetic field of a straight wire segment is Br) = µ I h cosα 1 cosα ) n 49) where the height h and the angles α 1 and α are as shown on figure below O r 1 h α 1 41) r r α and n is the unit vector to the whole triangle in the direction given by the right-screw rule. Note: in the limit of infinitely long segment in both directions, α 1, α π, hence cosα 1 cosα, and the magnetic field of the segment agrees with the formula for an infinite wire, B = µ I πh. 5) 11

12 Example#4: A Square Loop Consider a closed loop of wire in the shape of an a a square: ) Let s calculate the magnetic field at the center of the square shown in blue). The square wire consists of 4 similar straight-line segments, so all we need is to evaluate eq. 49) for the magnetic field due to each segment, and then total up the 4 segments contributions. For each segment, h = 1 a, α 1 = 45, α = 135, hence B 1segment = µ I a/) cos45 cos135 = ) = π µ I a. 5) Also, for each segment the triangle spanning the wire and the center of the square where we measure B lies in the plane of the square, so the direction of the magnetic field due to each segment is to the whole square. Specifically, the magnetic field points into the page since in each segment the current flows clockwise around the center. Thus, altogether, the magnetic field points into the screen and its magnitude is Example#5: Symmetric N-sided Polygon B whole square = 4 B 1segment = π µ I a. 53) In this example the wire also makes a complete loop, this time in the shape of symmetric N sided polygon with side a, for example 54) Again, we focus on the magnetic field at the center of the polygon, so by symmetry each 1

13 segment of the wire contributes a similar B 1segment. All these contributions are directed to the polygon, specifically into the screen, hence B polygon = N B 1segment n 55) where n is the unit vector pointing into the page. Now let s draw a single segment of the wire and the triangle connecting it to the center point where the magnetic field is measured: α 1 α h π N Simple geometry+trigonometry for this triangle gives us α 1, = π π N, 56) cosα 1 cosα = sin π N, 57) h = a ctan π N, 58) and therefore B 1segment = µ I cosα 1 cosα h = µ I tan N π a sin π N. 59) Finally, combining all N segments, we find the magnetic field at the center of the polygon is B = N B 1segment = N µ I a sin π N tan π N = µ I perimeter = Na N π sin π N tan π N. 6) To check this formula, we first plug in N = 4 for the square and compare with eq. 53) 13

14 from the previous example: for N = 4, N π sin π N tan π N = 16 π 1 = 8 π, 61) hence B[from eq. 6)] = µ I 4a 8 π = π µ I a, 6) in complete agreement with eq. 53) for the square. Second, let s take a large N limit in which the polygon becomes a circular ring of perimeter Na = πr. In this limit, N lim N π sin π N tan π ) N = π, 63) hence the magnetic field at the center of the polygon becomes B[from eq. 6)] µ I πr π = µ I R, 64) which indeed agrees with the magnetic field 38) at the center of a circular ring. Example#6: A Circular Arc As our final example, let s calculate the magnetic field at the center of a circular arc. More generally, consider a wire comprised of a semicircle and two straight segments ϕ 65) C and calculate the magnetic field at point C at the center of the circular arc. 14

15 Note: besidesbeingatthecenterofthearc, thepointchappenstolieonthestraight-line extrapolations of the straight segments of the wire. Consequently, the two straight segments do not contribute to the magnetic field BC) at that point; instead, the entire field at point C comes from the circular arc segment only. Let s parametrize the arc segment by the angle φ from the point C; φ ranges from φ to φ +ϕ. In terms of φ, r = Rcosφˆx + Rsinφŷ, 66) r r = Rcosφˆx Rsinφŷ, 67) dr = R sinφˆx + cosφŷ)dφ, 68) hence in the numerator of the Biot Savart Laplace integral dr r r ) = R dφ sinφcosφˆx ˆx + ŷ ŷ = ) + sin φˆx ŷ = +ẑ) cos φŷ ˆx = ẑ) 69) = R dφ sin φ+cos φ)ẑ = R dφẑ. Note: the direction of this vector product is always vertically Up, to the plane of the ring, so the magnetic field s direction is going to be vertically Up. As to the denominator of the BSL formula, the whole circular arc is at constant distance r r R from the ring s center, so the denominator is a constant R 3. Altogether, arc dr r r ) r r 3 = φ +ϕ φ R ẑdφ R 3 = ϕ ẑ, 7) R so the magnetic field at point C is Note: the ϕ angle in this formula should be taken in radians. BC) = µ I ϕ ẑ. 71) R 15

16 Magnetic Fields of Thick Conductors The Biot Savart Laplace formula Br) = µ wire I dr r r r r 3 11) gives the magnetic field of a steady current flowing through a thin wire which may be approximated by an infinitely thin line, straight or curved. For such a wire, it does not matter how the current is distributed across the wire s cross-section, only the net current I enter the formula. But sometimes we have currents flowing through the volume of a conductor which is too thick to be approximated as a line. For such conductors, wire I dr becomes conductor s volume d 3 Vol Jr ) 7) where Jr ) is the current density at the point r. Consequently, the Biot Savart Laplace equation for such current densities generalizes to Br) = µ d 3 Vol Jr ) r r r r 3. 73) Likewise, for a steady current flowing along a conducting surface with density Kr ), the BSL equation becomes Br) = µ d A Kr ) r r r r 3. 74) Note: the volume current density J is defined as current per unit of area to the current, while the surface current density K is defined as current per unit of length to the current. As a vector, K must be tangent to the surface; for example, for the surface spanning the x,y) plane, we may have any K x and K y components, but the K z component must vanish. 16

17 Example#7: Infinite Flat Current Sheet As our final example, consider infinite flat current sheet i.e., a D conducting surface carrying a uniform surface current K. Let s choose our coordinates such that the current flows in +ŷ direction along the surface spanning the x,y) plane. Then the magnetic field at some generic point r = x,y,z) is given by eq. 74), specifically, Bx,y,z) = µ K plane dx dy ŷ r r ) r r 3. 75) In the numerator inside the integral here, r r = x x )ˆx + y y )ŷ + zẑ, 76) ŷ r r ) = x x ) ẑ) + y y ) + z+ˆx) = x x )ẑ + zˆx, 77) while in the denominator r r 3 = x x ) +y y ) +z ) 3/. 78) To simplify these expressions, let s change the integration variables x and y to the polar coordinates s, φ) centered at x, y), thus x = x + s cosφ, 79) y = y + s sinφ, 8) dx dy = sdsdφ, 81) ŷ r r ) = scosφẑ + zˆx, 8) r r 3 = s +z ) 3/. 83) Plugging all these formulae into eq. 75), we arrive at Bs,y,z) = µ K dss π dφ s cosφẑ + zˆx s +z ) 3/. 84) 17

18 Integrating over the polar angle φ, we immediately obtain π π π dφ s cosφẑ+zˆx ) = sẑ cosφdφ + zˆx dφ = sẑ + zˆx π = πzˆx, 85) hence Bx,y,z) = µ K ˆx zsds s +z. 86) ) 3/ Note: the magnetic field everywhere points in either +ˆx or ˆx direction, depending on the sign of the remaining integral in this formula. To evaluate this integral, we change variables from s to t = s +z, thus zsds = z dt, z +s ) 3/ = t 3/, 87) hence zsds s +z ) 3/ = z z dt t 3/ = z [ ] + = z [ t z ] z = z z = signz), 88) and therefore Bx,y,z) = µ K signz) ˆx 89) Note: the magnetic field of the infinite current sheet is completely uniform above the sheet z > ) and likewise completely uniform below the sheet z < ), but at the current sheet itself z = ) there is a discintinuous jump. Its magnitude B of the field is the same µ K/ above and below the sheet, but the directions are opposite: above the sheet, the magnetic field points in the +ˆx direction while above the sheet it points in the ˆx direction. Relative to the currents direction +ŷ, the magnetic field above the sheet points 9 to the right of the current, while below the sheet it points 9 to the left of the current. 18

19 In a later lecture we shall learn that the discontinuity of the magnetic field across the current sheet discb) = µ K n 9) follows from the Ampere Law, just like the discontinuity of the electric field across a charged surface disce) = σ ǫ n 91) follows from the Gauss Law. However, the discontinuities of the electric and the magnetic fields have different directions: while the electric discontinuity disce) is to the whole charged surface, the magnetic discontinuity discb) is tangent to the current sheet but to the current s direction. 19